Proposition 10

Let $M$ be an invertible sub-A*-module & $B$. Then: (i) There exists $s \in S$ such that $As \subset M \subset As^{-1}$ (and in particular $M$ is non-degenerate). (ii) $A : M$ is the only sub-A*-module* $N$ of $B$ such that $M.N = A$.

If $M.N = A$, then $B.M = B.(B.M) \supset B.(M.N) = B.A = B$, hence $M$ is non-degenerate. Similarly $N$ is non-degenerate. If $t \in S \cap M$ and $u \in S \cap N$ (no. 5, Proposition 8 ), the element $s = tu$ belongs to $S \cap M \cap N$, whence $Ms \subset M.N = A$ and therefore $As \subset M \subset As^{-1}$. On the other hand, obviously $N \subset A : M$, whence

and $M.(A : M) = A$; multiplying the two sides by $N$, we deduce $A : M = N$, which completes the proof.